I am currently reading a book called Physics for Scientists and Engineers by Serway. While reading the chapter about 2-dimensional kinematics, I asked myself a question that I can't solve. The problem in the book is as follows:

A particle is moving in a plane, starting from (0,0). $V_{0x}=20m/s$, $A_x=4m/s$, $V_{0y}=-15m/s.$ Find a,b,c,d.....

From this data I wrote a displacement equation as follow:

$$\vec r = (20t+2(t^2))î - (15t)j $$

Then I asked if it is possible to find the distance traveled at a certain point in time (t=x), and not only the displacement. I couldn't seem to figure out how to do that! Thanks for any help.


2 Answers 2


Length: $\int_{t_0}^{t_1}| \vec{r} '(t)|dt$. In your case: $ \int_{0}^{t}\sqrt{(20+4t')^2+(15)^2}dt'$ Use Wolframalpha if you are now sure how to deal with it: https://www.wolframalpha.com/input/?i=int+sqrt(%2820%2B4x%29^2%2B%2815%29^2)dx

  • $\begingroup$ This is the integral of the magnitude of velocity, isn't it? It makes sense that Integrating the displacement vector as well will result in the distance traveled, because over tiny distances displacement and distance traveled are the same. Aren't they? $\endgroup$
    – rel-s
    Commented Aug 12, 2015 at 19:14
  • $\begingroup$ Exactly. $|\vec r'(t)|dt$ can (and should) be interpreted as the infinitesimal distance travelled between the times $t$ and $t+dt$. $\endgroup$
    – Antitheos
    Commented Aug 12, 2015 at 19:17

Since this is a specific question, I can't answer it directly for you but I may give you a hint: you already have the displacement formula $\stackrel{\to }{r}\left(t\right)$. From there you may find the relation y(x) using t as a parameter and use the formula for the arc length from calculus: $L=\underset{a}{\overset{b}{\int }}\sqrt{1+{\left({y}^{\prime }\right)}^{2}}dx$. Cheers!


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